solve the system (x^3)(y^3)(z^4)=1 and (x^2)(y^4)(z^4)=2 and (x^2)(y^3)(z^5)=3

Saturday, December 12, 2015

$\displaystyle{\left({{x}}^{{3}}\right)}{\left({{y}}^{{3}}\right)}{\left({{z}}^{{4}}\right)}=$
1................(1)

$\displaystyle{\left({{x}}^{{2}}\right)}{\left({{y}}^{{4}}\right)}{\left({{z}}^{{4}}\right)}=$
2..................(2)

$\displaystyle{\left({{x}}^{{2}}\right)}{\left({{y}}^{{3}}\right)}{\left({{z}}^{{5}}\right)}={3}$
....................(3)

First, we will divide (1) by
(2).

==> $\displaystyle\frac{{x}}{{y}}=\frac{{1}}{{2}}=\Rightarrow{y}={2}{x}$
..................(4)

Now we will divide (1) by
(3).

$\displaystyle=\Rightarrow\frac{{x}}{{z}}=\frac{{1}}{{3}}=\Rightarrow{z}=$
3x...................(5)

Now, we will substitute (5) and
(5) into (1).

$\displaystyle=\Rightarrow{\left({{x}}^{{3}}\right)}{\left({{y}}^{{3}}\right)}{\left({{z}}^{{4}}\right)}=$
1

$\displaystyle=\Rightarrow{\left({{x}}^{{3}}\right)}{{\left({2}{x}\right)}}^{{3}}{{\left({3}{x}\right)}}^{{4}}=$
1

$\displaystyle=\Rightarrow{\left({{x}}^{{3}}\right)}{\left({8}{{x}}^{{3}}\right)}{\left({81}{{x}}^{{4}}\right)}=$
1

$\displaystyle=\Rightarrow{648}{{x}}^{{10}}=$
1

$\displaystyle=\Rightarrow{{x}}^{{10}}=$
1/648

$\displaystyle=\Rightarrow{x}=$
(1/648)^(1/10)

$\displaystyle=\Rightarrow{y}={2}{x}=$
2/(648)^(1/10)

$\displaystyle=\Rightarrow{z}={3}{x}=\frac{{3}}{}$
648^(1/10)

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